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2 years ago

Find the surface area, including the floor, of his tent

Mathematics

1 answer:

Sholpan [36]2 years ago

8 0

Answer:

The surface area of the ten t including the floor is 52.8 m².

Step-by-step explanation:

We are given that,

Shaun's tent consists of 3 rectangles and 2 triangles.

1. The dimensions of the rectangles are 5 × 3 meters.

Since, Area of the rectangle = Length × Width

i.e. Area of 3 rectangles = 3 × Length × Width

i.e. Area of 3 rectangles = 3 × 5 × 3 = 45 m²

Thus, area of the 3 rectangles is 45 m².

2. The dimensions of the triangles are base= 3 meter and height= 2.6 meter

As, Area of the triangle =

i.e. Area of 2 triangles =

i.e. Area of 2 triangles = 7.8 m²

Thus, area of 2 triangles is 7.8 m².

So, the total surface area of the prism = Area of rectangles + Area of triangles

i.e. Total surface area = 45 + 7.8 = 52.8 m²

Hence, the surface area of the ten t including the floor is 52.8 m².Step-by-step explanation:

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5. One canned juice drink is 15% orange juice; another is 10% orange juice. How many liters of each should be mixed together in

hram777 [196]

X: liters of 15% orange juice
y: liters of 10% orange juice

x+y=5L

y=5L - x

x liters of 15% orange juice plus y liters of 10% orange juice makes 5 liters of 12% orange juice in equation form:

0.15x+0.10y = 0.12(5L)

0.15x+0.10(5L-x)=0.12(5L)

0.15x+0.5-0.10x=0.6

0.05x=0.1L

x=2L

so y = 3L

so you need 2 liters of 15% orange juice and 3 liters of 10% orange juice

4 0

3 years ago

The Ribys spent 5625 for 4.5 acres of land Find the price per acre

Margarita [4]

We can find the price per arc by dividing the total price by the amount of arc.
so in this case
5625 is the total money spent
and 4.5 is what they get when spending that amount of money
5625 / 4.5
= 1250
The price per arc is $1250
therefore, the answer is $1250.

4 0

3 years ago

Read 2 more answers

Which of the following options have the same value as 30\%30%30, percent of 818181?

Nataly_w [17]

Answer:

Option B is correct = $0.3 \times 81$

Step-by-step explanation:

The complete question is: Which of the following options have the same value as 30% of 81?

Group of choices is:

(A) $\frac{30}{100}\times 81 \times 100$

(B) $0.3 \times 81$

(D) $\frac{3}{10}\times 81 \times 10$

(E) $30 \times 81$

Now, the expression given to us is 30% of 81.

Simplifying the above expression we get;

30% of 81 = $\frac{30}{100} \times 81$

= $\frac{3}{10} \times 81$ = $0.3 \times 81$

Now, we will solve each of the given options and then see which option matches with our calculation.

Option (A) is given;

$\frac{30}{100}\times 81 \times 100$ = $30 \times 81$

This doesn't match with our answer, so this option is not correct.

Option (B) is given;

$0.3 \times 81$

This matches with our answer, so this option is correct.

Option (C) is given;

$0.03 \times 81$

This doesn't match with our answer, so this option is not correct.

Option (D) is given;

$\frac{3}{10}\times 81 \times 10$ = $3 \times 81$

This doesn't match with our answer, so this option is not correct.

Option (E) is given;

$30 \times 81$

This doesn't match with our answer, so this option is not correct.

6 0

3 years ago

Graph the first six terms of a sequence where a1 = 2 and r = 4.

zavuch27 [327]

We are given with
a1 = 2
r = 4

These are components of a geometric series. The first term is 2 and the common ratio is 4. To get the first six terms, we use the formula:
an = a1 r^(n-1)
a1 = 2 (4)^(1-1) = 2
a2 = 2 (4)^(2-1) = 8
a3 = 2 (4)^(3-1) = 32
a4 = 2 (4)^(4-1) = 128
a5 = 2 (4)^(5-1) = 512
a6 = 2 (4)^(6-1) = 2048

8 0

3 years ago

Read 2 more answers

In circle M, segment AB is tangent to the circle at point C. AB has endpoints such that AM BM  ,

iren2701 [21]

Answer:

Step-by-step explanation:

We can use the geometric mean theorem:

The altitude on the hypotenuse is the geometric mean of the two segments it creates.

In your triangle, the altitude is the radius CM and the segments are AC and BC.

$CM = \sqrt{AC \times BC} = \sqrt{ 9 \times 4} = \sqrt{36} = \mathbf{6}\\\text{The radius of the circle M is $\large \boxed{\mathbf{6}}$}$